When the discriminant is greater than zero, there are two real solutions.

When the discriminant is equal to zero, there is one real solution (a repeated root of multiplicity two).

When the discriminant is less than zero, there are two complex conjugate roots (no real solution).

As subhee has shown, the discriminant may be expressed as -(3p + 1)(5p - 1) which has zeroes of -1/3 and 1/5. So plot those points on a number line, creating the open intervals (where we may choose a test point from each interval):

(-?,-1/3) test point -1: -(-)(-) = -, no real solutions on this interval